Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**zetafunc.****Guest**

Hi, I'm a bit stuck on this problem. I am using an algebraic approach but maybe a geometric approach would be more efficient, I do not know.

"Let a, b and c be the lengths of the sides of a triangle. Suppose that ab + bc + ca = 1. Show that (a+1)(b+1)(c+1) < 4."

I used the arithmetic/harmonic mean inequality to get:

9abc ≤ a + b + c

But I don't really think this is useful. Can anyone give me a push in the right direction?

Thanks.

**zetafunc.****Guest**

Substitutions into that inequality also yield things like

but I still can't see how to use that.

**zetafunc.****Guest**

I also have

but I still can't see where this is going. Am I going to have to draw a picture of this at some point? Obviously the triangle is equilateral for the case a = b = c, but the objective of this problem does not seem to be concerned with specific cases.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,710

Hi;

Expanding out and using the constraint gets:

But I am not going anywhere quick from here.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**zetafunc.****Guest**

I noticed that too... but I am avoiding the temptation to try to work backwards. I mustn't try to show that if (a+1)(b+1)(c+1) < 4, then ab + bc + ca = 1.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,710

This is an Olympiad problem but I do not remember it and I did not write down the solution.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**zetafunc.****Guest**

Yes, it is from BMO Round 1 2010. But unfortunately the solutions are not available online... well, they are available, but you have to pay a lot of money for them.

**zetafunc.****Guest**

I have noticed that

But, I am not sure what to do from here.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,710

Hi zetafunc.;

Are you sure of that?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**zetafunc.****Guest**

Hmm, it looks like it isn't. I didn't check properly. I expanded the LHS by hand then used WolframAlpha to expand the RHS because I was lazy, and they don't match. darn.

**zetafunc.****Guest**

But,

so maybe it is useful...

**zetafunc.****Guest**

Sorry, that was stupid. What I meant was:

which is definitely correct.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,710

Hi;

That is not checking out.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**zetafunc.****Guest**

Sorry again. Subtract 2 from the LHS. That does it for sure. Then, noting that -[2(ab + bc + ca) + 2] = -4...

**zetafunc.****Guest**

Wait, but then I get the same thing in post #8 if you replace the 2 with 2(ab + bc + ca). I'm confused.

**zetafunc.****Guest**

Wait, surely this problem is solved then? Because clearly (a-1)(b-1)(c-1) is smaller than zero, so set that LHS to less than 0 and you get their inequality... is that a valid solution?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,710

There was a constant in there that you have left out.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**zetafunc.****Guest**

Where?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,710

In my feeble brain. It appears to be gone now.

I have:

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**zetafunc.****Guest**

But

Clearly,

so,

and therefore

or am I wrong?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,710

Hi;

Your second line, why is that less than 0?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**zetafunc.****Guest**

Because ab + bc + ca = 1, so a, b and c are all smaller than 1, and therefore each term (a-1), (b-1) and (c-1) is negative, so the product of those three terms if also negative and therefore smaller than zero.

**zetafunc.****Guest**

And also, a, b and c are all greater than 0 (they're sides of a triangle).

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,710

Because ab + bc + ca = 1, so a, b and c are all smaller than 1

Can you prove that mathematically?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**zetafunc.****Guest**

The equation factorises to these three:

a(b + c) + bc = 1

b(a + c) + ca = 1

c(a + b) + ab = 1

so surely a, b and c must all be smaller than 1?